posted by Anonymous .
Can someone check my answers for the following questions...btw HAPPY HOLIDAYS TO EVERY AT JISKHA!
1)As the magnitude of Keq increases, the amount of product formed in a reversible reaction: REMAINS CONSTANT
2)The Keq of reacion 4 x 10-7. At equilibrium: THE REACTANTS ARE FAVORED
3)To increase the yield of CH3OH for the equilibrium reaction CO(g) + 2H2 (g) <--> CH3OH (g) + 24 kcal: INCREASE THE T AND DECREASE THE P
1) Consider again the definition of Keq. The concentrations of products appear in the numerator, and the concentrations of reactancs appear in the numerator. If Keq increases, the concentrations of products must increase.
3) Apply Le Chatelier's principle here. The forward reaction is exothermic and it also decreases the number of moles in a fixed volume. To increase yield, you want to favor the forward reaction. Increasing temperature and decreasing pressure favors the REVERSE reaction, by tending to compensate for the temperature rise (absorbing heat) and pressure decrease (producing more moles)